I very much think the other tutor completely missed the point of your question....
Yes, you can "multiply out" the expression ((x+5)^(0.5)) using binomial expansion.
Many students will be familiar with binomial expansion when the power is an integer; few students will be familiar with it when the power is not an integer.
Because the exponent is not an integer, the expansion does not terminate; it has an infinite number of terms. The first few terms will give good approximations.
The key to understanding binomial expansion with non-integer powers is to look at how binomial coefficients are calculated.
For all n, including non-integer values of n...
For all other binomial coefficients C(n,r), the coefficient can be calculated as a fraction with "r" numbers in both numerator and denominator, starting with n and decrementing by 1 in the numerator, and starting with r and decrementing by 1 in the denominator. That gives us...
etc....
For example for n=3, we get what (I hope) are the familiar coefficients for n=3:
And using the same method for calculating the coefficients for n=0.5, we get the unfamiliar coefficients for a binomial expansion to the 0.5 power:
Now we can write out the first few terms of the binomial expansion of (x+5)^(-0.5). We have the binomial coefficients; and from one term to the next the power of x decreases by 1 and the power of 5 increases by 1:
Evaluating that expression for values of x that should produce whole number results, we find
x x+5 (x+5)^(0.5) f(x) (4 decimal places)
------------------------------
4 9 3 3.1035
11 16 4 4.0042
20 25 5 5.0006
31 36 6 6.0001
44 49 7 7.0000
59 64 8 8.0000
For small values of x, where this approximation is not very accurate, using 2 or 3 more terms of the binomial expansion would improve the accuracy.